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How do we know that wave f. is the eigenfunction of an operator H?

  1. May 10, 2013 #1
    I am kind of new to this eigenvalue, eigenfunction and operator things, but i have come across this quote many times:
    First i need some explaination on how do we know this? All i know about operator ##\hat{H}## so far is this equation where ##\langle W \rangle## is an energy expected value:

    \begin{align}
    \langle W \rangle &= \int \limits_{-\infty}^{\infty} \overline{\Psi}\, \left(- \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p\right) \Psi \, d x
    \end{align}

    From which it follows that ##\hat{H} = - \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p##.

    Aditional question:

    I know how to derive relation ##\hat{H}\hat{a} = (W - \hbar \omega)\hat{a} \psi## for which they state that:

    I also know how to derive relation ##\hat{H}\hat{a}^\dagger = (W + \hbar \omega)\hat{a}^\dagger \psi## for which they state that:

    How do we know this?
     
  2. jcsd
  3. May 10, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    That's pretty much the definition of "eigenfunction" and "eigenvalue", isn't it?
     
  4. May 10, 2013 #3
    What exactly here is the definition? Which equation?
     
  5. May 10, 2013 #4
    Any equation of the form
    ##\hat{H} \psi = E \psi##
    is an eigen-equation

    The functions ##\psi_n## which satisfy this equation are called the eigenfunctions of the operator ##\hat{H}##, with corresponding eigenvalues ##E_n##.

    Put simply, when an operator ##\hat{H}## acts upon one of its eigenfunctions, it returns the same eigenfunction multiplied by a number, which we call the eigenvalue.
     
  6. May 11, 2013 #5
    1st:
    In this equation ##\hat{H}\hat{a} = (W - \hbar \omega)\hat{a} \psi## is it possible that ##\hat{a} \psi## is an eigenfunction... i mean ##\hat{a}## is an operator...

    2nd:
    And is it possible that my quotings were wrong i will write the wrong one and the one that i think is better:

    WRONG:
    ##\hat{a}\psi## is an eigenfunction of operator ##\hat{H}## with eigenvalue ##(W−\hbar \omega)##.

    RIGHT:
    ##\hat{a}\psi## is an eigenfunction of operators ##\hat{H}\hat{a}## with eigenvalue ##(W−\hbar \omega)##.
     
  7. May 11, 2013 #6

    Nugatory

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    Staff: Mentor

    What's right is wrong and what's wrong is right :smile:
    An operator maps one function into another, so if ##\hat{a}## is an operator and ##\psi## is a function, then ##\hat{a}\psi## is itself a function.

    And I think that you dropped a ##\psi## from the equation you started with. I would have expected ##\hat{H}\hat{a}\psi = (W - \hbar \omega)\hat{a} \psi##, and written that way it is clear that the function ##\hat{a}\psi## is an eigenfunction of ##\hat{H}## with eigenvalue ##(W−\hbar \omega)##... Or at least adding a few parentheses will make it clear:

    [tex]\hat{H}(\hat{a}\psi) = (W - \hbar \omega)(\hat{a} \psi)[/tex]
     
  8. May 12, 2013 #7
    Ok so the eigenfunction can be a combination of an operator $\hat{a}$ and a function $\psi$?
     
  9. May 12, 2013 #8
    Why not? An operator acting on a function gives you another function.
     
  10. May 12, 2013 #9
    Thanks.
     
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